resistorsvoltage
We did an experiment using a very simple circuit shown below:
The voltage drops over each resistors were: 4.9, 2.8, and 2.2 respectively. According to the rules the voltage drops across each resistor must add up to the voltage of the source, which in this case was 5 V, however this is not the case. Why is that?
The LED forward voltage drop will remain (roughly) the same, but the current can change, so the calculation becomes (same equation solving for I):
$$I_{LED} = {(V_s - V_f)\over{R}}$$
So for a 3V \${V_f}\$ and a 5V supply, the \$100\Omega\$ resistor would give \${(5V - 3V)\over{100 \Omega }} = 20 mA\$.
So if you know what current you want, just plug the values in, e.g. for 10mA:
$$R = {(5V - 3V)\over{0.01 A}} = {200 \Omega}$$
Basically, the fact that the supply and the LED forward voltage can be relied upon to be pretty static, means that whatever value resistor you put in will also have a static voltage across it (e.g ~2V in this case), so it just leaves you to find out that voltage and select a resistance value according to the current you want.
Below is the V-I curve of a diode (from the wiki LED page), notice the current sharply rises (exponentially) but voltage stays roughly the same when the "on" voltage is reached.
For more accurate control of the current you would use a constant current, which is what most LED driver ICs provide.
Sounds to me like you might have the regulator wired wrong. Double check that it is wired correctly:
Capacitors are optional according to most datasheets, but are good practice to include.
Best Answer
simulate this circuit – Schematic created using CircuitLab
Figure 1. Measuring voltage across each resistor. Figure 2. Measuring voltages at various points in a circuit using a common reference.
Make sure you are measuring at shown in Figure 1. Alternatively use the method of Figure 2 and subtract voltages to calculate the voltage across each resistor.